chi-square distribution >> gamma(n/2)

Question

My professor showed the transformation from chi-square to gamma(n/2), but I don't understand it.

Let X be the chi-square distribution with m degrees of freedom. If Y=X/2, Y becomes gamma(n/2). What is the proof underlying this result??

Answer 1

A Gamma distribution is the distribution of the sum of independent exponentials, whereas the chi squared distribution is the distribution of the sum of squares of independent normals.

The distribution of the sum of squares of two standard normals, i.e. a $\chi^2(2),$ can be computed by writing it as $Z = X^2+Y^2$ and then using polar coordinates: $$ P(Z \le z) = P(X^2 + Y^2 \le z) = \iint_{x^2+y^2\le z} \frac{1}{2\pi}e^{-\frac{1}{2}(x^2+y^2)}\,dxdy=\int_0^{\sqrt{z}}e^{-\frac{1}{2}r^2}r\,dr =1-e^{-\frac{1}{2}z}.$$ So we see that a $\chi^2(2)$ is an exponential with mean $2.$ So if $X\sim\chi^2(2)$ then $Y=X/2$ is a standard exponential, which is a $Gamma(1).$ So this proves your identity for $n=1.$

Then for $n=2$, a $Gamma(2)$ is the sum of two independent standard exponentials. As we showed before, each of those exponentials are equal in distribution to half of the sum of squares of two standard normals, so the $Gamma(2)$ is equal in distribution to half of the sum of squares of four independent normals, i.e. to one half of a $\chi^2(4).$

Similarly for any integer.